How to evaluate the closed form of $I=\int_{0}^{1}\ln[\sin(\pi x)]\ln^2[2\sin(\pi x)]\mathrm dx?$

Question

Evaluate the closed from of:

$$I=\int_{0}^{1}\ln[\sin(\pi x)]\ln^2[2\sin(\pi x)]\mathrm dx$$

We have the following:

$$\int_{0}^{1}\ln^2[2\sin(\pi x)]\mathrm dx={\zeta(2)\over 2}\tag2$$

$$\int_{0}^{1}\ln[\sin(\pi x)]\mathrm dx=-\ln(2)\tag3$$

Approximately

$$I\approx-{\zeta(2)\over \ln (2)}$$

Indefinite integral:

$u=\sin(\pi x)\implies du=\pi\cos(\pi x) dx$

$${1\over \pi}\int{\ln(u)\ln^2(2u)\over \sqrt{1-u^2}}\mathrm du\tag4$$

$$(1-x)^{-1/2}=\sum_{k=0}^{\infty}{(2k-1)!!\over (2k)!!}x^k$$

$${1\over \pi}\sum_{k=0}^{\infty}{(2k-1)!!\over (2k)!!}\int{ u^{2k} \ln(u)\ln^2(2u)}\mathrm du\tag5$$

Answer 1

Hint. One may recall the Euler beta function, giving

$$ \int_0^{\pi}(\sin\theta)^{a}\,d\theta=2\cdot\frac{\Gamma(\frac12)\,\Gamma(\frac {a+1}2)}{\Gamma(\frac12+\frac {a+1}2)}=\sqrt{\pi}\cdot\frac{\Gamma(\frac {a+1}2)}{\Gamma(\frac {a}2+1)},\qquad a>-1, $$ or

$$ \int_0^{1}(2\sin \pi x)^{a}\,dx=\frac{2^a}{\sqrt{\pi}}\cdot\frac{\Gamma(\frac {a+1}2)}{\Gamma(\frac {a}2+1)},\qquad a>-1,\tag1 $$

then by differentiating with respect to $a$, observing that $$ \frac{d^2}{da^2}\left.(2\sin \pi x)^{a}\right|_{a=0}=\ln^2(2\sin \pi x), \quad \frac{d^3}{da^3}\left.(2\sin \pi x)^{a}\right|_{a=0}=\ln^3(2\sin \pi x), $$ and observing that $$ \frac{d^2}{da^2}\left(\left.\frac{2^a}{\sqrt{\pi}}\cdot\frac{\Gamma(\frac {a+1}2)}{\Gamma(\frac {a}2+1)}\right)\right|_{a=0}=\frac{\pi^2}{12}, \quad\frac{d^3}{da^3}\left(\left.\frac{2^a}{\sqrt{\pi}}\cdot\frac{\Gamma(\frac {a+1}2)}{\Gamma(\frac {a}2+1)}\right)\right|_{a=0}=-\frac32\cdot \zeta(3), $$ one gets

$$ I=\int_{0}^{1}\ln[\sin(\pi x)]\ln^2[2\sin(\pi x)]\mathrm dx=\color{blue}{-\frac{\pi^2}{12}\cdot \ln 2-\frac32\cdot \zeta(3).} \tag2 $$